At 9:00 on Saturday morning, two bicyclists heading in opposite directions pass each other on a bicycle path.The bicyclist heading north is riding 6 km/hour faster than the bicyclist heading south. At 10:15, they are 42.5 km apart. What are their rates?

1 Answer
Jun 17, 2017

#"North bicycle"=20" "(km)/(hour)#

#"South bicycle"=14" "(km)/(hour)#

Explanation:

We need to set up equations based on the known information. I always write down everything the question gives me and try to link the information with logic or established formulas. I automatically think of speed=distance/time in this case. Also, units are important, so decide what you want to use and stick with them for all calculations. I will use km and hours.

Okies, the total distance travelled by both cyclists is 42.5 km. This means that the sum of the distances will be 42.5 km. Let's use #d_N# as the distance travelled by the cyclist going north, and #d_S# for the cyclist going south:

#d_N+d_S=42.5" "km# (1)

Rearrange the speed formula to give the distance travelled as speed multiplied by time. We will use #N# for the speed of the northward bound cyclist, and #S# for the other cyclist. The time is from 9:00 to 10:15, so t = 1.25 hours. Now, we can write expressions for the distance travelled by each cyclist:

#d_S=1.25S# (2)

Remember that the northward bound cyclist is going 6 km/hour faster, so:

#N=S+6#

#rArrd_N=1.25N=1.25(S+6)=1.25S+7.5# (3)

At this point, we want to solve equation (1), so we need to replace either #d_N# with #d_S#, or vice versa, so that there is only one variable to solve. This means we need a second expression. We do this by evaluating (3) - (2):

#d_N=1.25S+7.5# (3)
- #d_S=1.25S# (2)

#rArrd_N-d_S=1.25S-1.25S+7.5#

#rArrd_N-d_S=7.5rArrd_N=d_S+7.5# (4)

Now substitute (4) into (1) to solve for #d_S#:

#d_N+d_S=42.5rArr(d_S+7.5)+d_S=42.5#

#rArr2d_S+7.5=42.5rArr2d_S=35rArrd_S=17.5" "km#

#d_N=42.5-17.5=25" "km#

Now we know the distances travelled by both so we use speed = distance/time to get the rates:

#N=25/1.25=20" "(km)/(hour)#

#S=17.5/1.25=14" "(km)/(hour)#